YES

We show the termination of the TRS R:

  f(|0|(),|1|(),x) -> f(s(x),x,x)
  f(x,y,s(z)) -> s(f(|0|(),|1|(),z))

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(|0|(),|1|(),x) -> f#(s(x),x,x)
p2: f#(x,y,s(z)) -> f#(|0|(),|1|(),z)

and R consists of:

r1: f(|0|(),|1|(),x) -> f(s(x),x,x)
r2: f(x,y,s(z)) -> s(f(|0|(),|1|(),z))

The estimated dependency graph contains the following SCCs:

  {p1, p2}


-- Reduction pair.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(|0|(),|1|(),x) -> f#(s(x),x,x)
p2: f#(x,y,s(z)) -> f#(|0|(),|1|(),z)

and R consists of:

r1: f(|0|(),|1|(),x) -> f(s(x),x,x)
r2: f(x,y,s(z)) -> s(f(|0|(),|1|(),z))

The set of usable rules consists of

  (no rules)

Take the reduction pair:

  lexicographic combination of reduction pairs:
  
    1. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2,x3) = max{0, x1 - 4, x3 - 1}
          |0|_A = 5
          |1|_A = 4
          s_A(x1) = x1 + 3
    
      precedence:
      
        f# = |0| = |1| = s
      
      partial status:
    
        pi(f#) = []
        pi(|0|) = []
        pi(|1|) = []
        pi(s) = []
    
    2. weighted path order
    
      base order:
    
        max/plus interpretations on natural numbers:
        
          f#_A(x1,x2,x3) = 0
          |0|_A = 2
          |1|_A = 5
          s_A(x1) = x1 + 6
    
      precedence:
      
        s > |0| = |1| > f#
      
      partial status:
    
        pi(f#) = []
        pi(|0|) = []
        pi(|1|) = []
        pi(s) = []
    

The next rules are strictly ordered:

  p2

We remove them from the problem.

-- SCC decomposition.

Consider the dependency pair problem (P, R), where P consists of

p1: f#(|0|(),|1|(),x) -> f#(s(x),x,x)

and R consists of:

r1: f(|0|(),|1|(),x) -> f(s(x),x,x)
r2: f(x,y,s(z)) -> s(f(|0|(),|1|(),z))

The estimated dependency graph contains the following SCCs:

  (no SCCs)